import java.util.LinkedList;
import java.util.Queue;

public class MyBinaryTree {

    static class TreeNode {

        public char val;

        public TreeNode left;//左孩子的引用

        public TreeNode right;//右孩子的引用


        public TreeNode(char val) {

            this.val = val;

        }

    }


    /**
     * 创建一棵二叉树 返回这棵树的根节点
     * 先使用枚举的方式来创建
     *
     * @return
     */

    public TreeNode createTree() {
        TreeNode A = new TreeNode('A');
        TreeNode B = new TreeNode('B');
        TreeNode C = new TreeNode('C');
        TreeNode D = new TreeNode('D');
        TreeNode E = new TreeNode('E');
        TreeNode F = new TreeNode('F');
        TreeNode G = new TreeNode('G');
        TreeNode H = new TreeNode('H');

        A.left = B;
        A.right = C;
        B.left = D;
        B.right = E;
        C.left = F;
        C.right = G;
        E.right = H;
        return A;

    }


    // 前序遍历

    public void preOrder(TreeNode root) {
        if (root == null) return;
        System.out.print(root.val + " ");
        preOrder(root.left);
        preOrder(root.right);
    }


    // 中序遍历

    void inOrder(TreeNode root) {
        if (root == null) return;
        inOrder(root.left);
        System.out.print(root.val + " ");
        inOrder(root.right);
    }


    // 后序遍历

    void postOrder(TreeNode root) {
        if (root == null) return;
        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val + " ");
    }


    public static int nodeSize;


    /**
     * 获取树中节点的个数：遍历思路
     */

    void size(TreeNode root) {
        //先序遍历思想
        if (root == null) return;
        nodeSize++;
        size(root.left);
        size(root.right);
    }


    /**
     * 获取节点的个数：子问题的思路
     *
     * @param root
     * @return
     */

    int size2(TreeNode root) {
        if (root == null) return 0;
        return size2(root.left) + size2(root.right) + 1;
    }

    /*
     获取叶子节点的个数：遍历思路
     */

    public static int leafSize = 0;


    void getLeafNodeCount1(TreeNode root) {
        if (root == null) return;
        if (root.left != null && root.right != null) {
            leafSize++;
        }
        getLeafNodeCount1(root.left);
        getLeafNodeCount1(root.right);
    }



    /*
     获取叶子节点的个数：子问题
     */

    int getLeafNodeCount2(TreeNode root) {
        if (root == null) return 0;
        if (root.left == null && root.right == null) {
            return 1;
        }
        return getLeafNodeCount2(root.left) + getLeafNodeCount2(root.right);
    }



    /*
    获取第K层节点的个数
     */

    int getKLevelNodeCount(TreeNode root, int k) {
        if (root == null) {
            return 0;
        }
        if (k == 1) {
            return 1;
        }
        return getKLevelNodeCount(root.left, k - 1) + getKLevelNodeCount(root.right, k - 1);
    }



    /*

     获取二叉树的高度

     时间复杂度：O(N)

     */

    int getHeight(TreeNode root) {
        if (root == null) return 0;
        int leftH = getHeight(root.left);
        int rightH = getHeight(root.right);
        return leftH > rightH ? leftH + 1 : rightH + 1;
    }


    // 检测值为value的元素是否存在

    TreeNode find(TreeNode root, char val) {
        if (root == null) return null;
        if (root.val == val) return root;

        TreeNode ret = find(root.left, val);
        if (ret != null) {
            return ret;
        }

        ret = find(root.right, val);
        if (ret != null) {
            return ret;
        }

        return null;
    }


    //层序遍历

    void levelOrder(TreeNode root) {
        if (root == null) return;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            System.out.println(node.val + " ");
            if (node.left != null) {
                queue.offer(node.left);
            }
            if (node.right != null) {
                queue.offer(node.right);
            }
        }
        System.out.println();
    }


    // 判断一棵树是不是完全二叉树

    boolean isCompleteTree(TreeNode root) {
        //利用层次遍历的思想将所有的节点入队 包括空孩子节点
        if (root == null) return false;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if (cur != null) {
                queue.offer(cur.left);
                queue.offer(cur.right);
            } else {
                break;
            }
        }
        //再次遍历队列 判断队列当中是否存在非空的元素
        while (!queue.isEmpty()) {
            TreeNode cur = queue.peek();
            if (cur == null) {
                queue.poll();
            } else {
                return false;
            }
        }
        return true;
    }
}
